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Chapter 7 Homogeneous Equations of the Second Degree
CHAPTER 7
Homogeneous Equations of the Second Degree 1. Let
be an indefinite quadratic form with integer coefficients and of determinant D = IIursll # 0. We discuss the solvability in integers of the equationf(f(m) = 0. Results for n = 3 were given by Legendre for the diagonal forms, i.e. a23 = a31 = a,, = 0, and various modifications of his proof have been given by other writers. Results for diagonal forms in four and five variables were found by Meyer nearly a hundred years after Legendre's work. The general case was considered by Minkowski and Hasse' culminating in the latter's result given below as Theorem 1 , and embodying a new and important principle, namely, that solvability of the equation f(x) = 0 for all local (i.e. p-adic) fields implied global solvability (i.e. solution in integers). Other results give estimates for the magnitude of a solution in terms of the coefficients. For n = 3 , the first sharp results are due to Holzer,. His method is rather advanced, and so further on we shall give the less sharp results due to Mordel13 which can be proved by elementary means. These were also found later by S k ~ l e mMordel15 .~ has since found a very elementary proof of Holzer's results also given further on. An estimate sometimes better than Holzer's and depending upon results in quadratic fields was given by M. Kneser ". Estimates for the magnitude of a solution of the general quadratic equation have been given by Cassels7, Davenport8, Birch and Davenport', but they have no place here. These results are of great importance in many investigations. Theorem 1 The equation f(x) = 0 is solvable in integers if and only if the congruence f(x) = 0 (modp') is solvable for all primes p and integers r 2 1 with (x1, x2,. . ., xn, P) = 1. Stated otherwise the conditions imposed uponf(s) mean that the equation f(5) = 0 is solvable in every p-adic field including of course the real field pa. The condition is obviously necessary. It is well known that by means of a linear substitution X, = c,,xr . . . c,,x,, (r = 1 , . . ., n),
+
+
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
where the c are integers and c,, # 0, that the solution of the equationf(.c) is equivalent to that of an equation F(%)
=
A,Xf
+
. ..
+ AnXZ = 0.
43 =
0
(2)
This is easily proved by induction. We may suppose that a,, # 0 by a linear substitution if need be. Then
=
aIIf(4
X ? f fib),
where Similarly = X;
h22f1(.7.)
where
+
S2(T),
2 b2rxr, etc. n
X,
=
r=2
It can be shown that the substitution does not affect p-adic solubility. 2. The case n = 2 is trivial and so we commence with n = 3 and write f ( x ,y , Z)
=
ax2
+ by2 + C Z ~= 0.
(3)
We now may suppose that a, 6, c have no common factor, and also that x, y , z have no common factor. We also suppose that a, b, c do not all have
the same sign and thatf(x, y , z ) 3 0 (mod p') is solvable for all primes p and integers r >, 1 with (x, y , z , p) = 1. We now reducef(x, y , z ) to a canonical form. Firstly, we may suppose that a, 6, c are square-free. For let a = U , C X ~b, = b,p2, c = c,y2 where a,, b,, c1 are square-free. We have now a 1-1 correspondence between the solutions of equation (3) and of
m,,z1)
=
y1,
alx? + b,Yf
+ clz? = 0
given by n,x, =
ax,
d.Y = pyx,,
cllyl
=
py,
fly
=
yay,,
d,z, dz
= yz,
= apz1,
(x1, y,, Zl) = 1,
( x , y , z ) = 1.
y , z ) = 0 is p-adically solvable so is f,(x, y , z ) = 0 and conClearly if ,/(s, versely. Next we may suppose in equation (3) that (6, c) = (c, a ) = (a, b) = 1. For let p be a prime factor of (b, c) and so b = pbl, c = p c l . Then ax2 = 0 (modp), and so ,Y = p x l . Hence with a , = pa, (3) becomes a,s:
+ b,y2 +
c,z2
=
0.
44
DIOPHANTINE EQUATIONS
Here alblcl = abc/p and so lalblcll< label. Clearly the process, applied to all the prime factors of (6, c), (c, a), (a, b), must terminate at a stage when (6, c) = (c, a) = (a, b) = 1. Further the final equation is solvable as a congruence for all p'. We now write equation (3) in the form
ax2 + by2 -
CZ'
=
0, u > 0, b > 0, c > 0.
(4)
Let p be an odd prime divisor of a. Then by2 - cz2 = 0 (mod a) is solvable and so as in Mordel13 and Skolem4, by2 - cz2 = b(y
+ kz)(y - kz) (modp),
f(x, Y , 4 = LlM1 (mod P), identically in x, y , z where L1,M 1are linear forms in x, y , z. Similarly if q, r are odd prime divisors of 6, c, respectively.
say, and so
f(x, y , 4 = L2M2 (mod 41, f(x, y , z ) = L3M3(mod r).
ax2 + by2 - cz2 = (ax + by
Also
+ C Z ) ~(mod 2).
Hence since abc is square-free, on taking into account the various p, q, r, we have by the Chinese remainder theorem, f ( x , y, z ) z L M (mod abc),
where L,M are linear forms. We show now that we can find integers x, y, z # 0, 0, 0, such that
ax2 + by2 - cz2 = 0 (mod abc),
and
0 < 1x1 <
dbc,
0 6 (yI <
d/ca,
0
< IzI
(5)
<
d/ab,
provided we exclude the trivial case a = b = c = 1. Here the numbers dbc, dG, dab are irrational. The number of sets of integers x,y, z with
o~(db abc. But L has only abc residues mod abc, and so for two sets of values for x,y , z, say, xl, yl,z1 and x2, y2, z2,
Uxi, yi, 21) Hence and
L(xl Ixl - x21 <
Then for x
=
- x2, yl
dbc,
2 z2) ,
(mod abc).
- y2, z1 - z2) = 0 (mod abc), Iyl - y21 < dG, Izl - z21 < dab.
x1 - x2,etc. -abc < axa
and so
L(x2,~
+ by2 -
ax2 + by2 -
CZ'
< bbc,
cz2 = 0 or abc.
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
45
In the second case,
+ by2)(z2 + ab) = c(z2 + ab)2, a(xz + by)2 + b(yz - ax)2 = c(z2 + ab)2. a x 2 + bY2 = cz2. (ax2
or
i.e.
This proves the solvability of equation (4). The result gives an estimate for the magnitude of the solution, but a better one is given by Mordell’s Theorem 2 There exists a non-triaial solution with 1x1 ,<
lyl <
d S C ,
d%,
IZI
< 2dab.
(6)
This is not so precise as Holzer’s2 result 1x1 < 6, (yl < di,IzJ < dab, where the equality signs can be omitted unless two of a, b, c, are equal to one. It suffices for a proof of this to show that we can find a solution not (0, 0,O) of the congruence f ( x , y , z) = ax2
with Then
1x1
< dZC,
+ by2 - cz2 = 0 (mod 4abc), Jyl <
dZa,
IZI
< 2dab.
If’(x, y , z)l < 4abc, i.e. f(x, y , z)
=
0.
We do this by applying Theorem 3 of Chapter 6 to five linear congruences. Suppose first that a, b, c are all odd. Then there exist integers A , B, C such that bA2 - c
= 0 (mod a), - cB2 + a = 0 (mod b),
aC2
+ b = 0 (mod c).
Three of the congruences are given by y - Az
= 0 (mod a),
z - Bx
= 0 (mod b),
x - Cy
= 0 (mod c).
Then from the first, ax2 + by2 - cz2 = (bA2 - c)z2 = 0 (mod a). Similarly for mod 6, mod c. We can satisfyf(x, y , z) = 0 (mod 4) by taking two linear congruences each to mod 2. We cannot have both a - c G 2 (mod 4), b - c = 2 (mod 4) for then a = b = - c (mod 4), and sof(x, y , z ) = 0 (mod 4) is not solvable. We may suppose that h - c = 0 (mod 41, and then the two congruences are x
= 0 (mod 2),
y
= z (mod 2).
46
DIOPHANTINE EQUATIONS
It now suffices to take rl
=
d/2bc,
r2
=
.\/2ca,
r3 = 2 4 2 .
in Theorem 3 of Chapter 6. Suppose next that one of a, b, c is even, say a. We can now make f(x, y , z ) = 0 (mod 3abc) if x, y , z satisfy the three congruences: y - Az
= O(mod+a),
z - Bx
= O(modb),
x - Cy
= O(modc).
We now satisfy f(x, y , z ) = 0 (mod 8) by taking two linear congruences one to mod 2 and the other to mod 4. These are x where I, m satisfy
+
= 1y (mod 2),
z
=
m y (mod 4),
a12 + b - em2 = 0 (mod 8).
If a b - c = 0 (mod 8), we take I = m = 1. If b - c = 0 (mod 8), we take I = 0, m = 1. Then as before, Theorem 3 of Chapter 6 gives a solution. The equation is impossible when b - c = 4 or a (mod 8 ) as follows on taking congruences mod 8. This finishes the proof. To summarize, we have proved
Theorem 3 rfa, b, care square free, (a, b) = (b, c) the same sign, then the equation ax2
=
(c, a )
=
1 , anda, b, c do not have
+ by2 + cz2 = 0
has non-trivial integer solutions if and only if - bc is a quadratic residue of a, and so for every prime factor of a, i.e. x2 + bc = 0 (mod a) is solvable, and similarly for - ca and - ab, and if ax2 + by2 + cz2 = 0 (mod 8)
is solvable. These give N , say, necessary and sufficient conditions. It can be shown that if any N - 1 of these conditions are satisfied, so is the remaining one. This is a consequence of the laws of quadratic reciprocity and the supplementary laws, namely, that if p , q are two different odd primes
and
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
47
There is, however, no need here to go into details. The result is well known from Hilbert’s theory of the norm-residue symbol. Theorem 4 If one integer solution not (0, 0,O)of the irreducible equation
f ( x ,y , z )
=
ax2
+ by2 + cz2 + 2fyz + 2gzx + 2hxy = 0
exists, then the general solution with ( x ,y , z ) expressions of the form x = Alp2
=
(7) 1 is given by a j k i t e number of
+ Bipq + Clq2, y = A2p2 + B 2 ~ 4+ C2q2, z = A3p2 + B3pq + c3q2,
(8) where p , q take all integer values with ( p , q) = 1, and the A , B, C are integer constants. For let a solution (x,, yo, z,) be known. Write z = rz,, x = rx, p, y = ry, q, where p , q, r are rational numbers. Then we have a linear equation in r from which r = f(p, q, O)/L(p,q ) where L is a linear function of p and q. Hence ignoring denominators, we have with an appropriate factor 8,
+
+
8 . ~= f l ( P , q),
8Y = f X P , q), 8,z = f 3 ( P , q), where fi,f2,f3 are binary quadratics with integer coefficients, and we may now suppose that p , q are integers and that ( p , q ) = 1. A simple elimination procedure shows that the common factors of f l ,f 2 , f 3are independent of p , q. These and S can be removed on replacing p , q by a linear substitution on P. 4. The method of Theorem 4 has been used by Mordel15to give an elementary proof of Holzer’s
Theorem 5 The solvable equation ax2 + by2 + cz2 = 0 taken in the canonicalform with a > 0, b > 0, c < 0 has a non-trivial solution with 1x1 I
dm,
dW,
IyI I
IZI
I
dZ
It is easily seen that all the equality signs may be removed unless two of the a, b, c are unity. Let (x,, yo, z,) be a solution with (x,, yo, z,) = 1. We show that if for this solution, ( z o ( > dab, then another solution (x, y , z) can be found with IzI < lzol. Hence there exists a solution ( x , y , z ) with IzI I dab. The inequalities for x , y are now obvious. Put
x
+ tx,
+
z = zo y = y, tY, where X , Y, 2, are integers to be determined later. Then = x,
(ax2
+ tz,
+ b Y 2 + cZ2)t + 2(axoX + byoY + czoZ) = 0.
48
DIOPHANTINE EQUATIONS
1
Hence, on neglecting a denominator, a solution, say, (x, y , z), is given by 6~ = zo(aX2
ax Sy
= =
xo(aX2 yo(aX2
+ 6 Y 2 + cZ') - 2Z(axoX + by0 Y + CZOZ), + b Y2 + - 2X(axoX + by, Y + C Z O Z ) , + 6 Y 2 + cZ') - 2 Y(uxOX + by0 Y + CZOZ), CZ2)
(9)
where S is a common divisor of the three expressions on the right. Further x , y , z are integers if
Ale,
6l(Xyo
For from ax;
-
Yxo),
+ by; + cz; = 0,
it easily follows that (S, abxoyo) = 1. From (9) it suffices to show that Q = a X 2 + b Y 2 = O(mod6). P = a x o X + 6yoY = O(mod6), Since also
X
= xo Y/yo, Q
E
P
(ax;
=
Y(axi
+ by;)/yo = 0 (mod 8);
+ by:) Y z / y ; E 0 (mod 6).
From (9)
Suppose now that z i > ab. Take X , Y as a solution of y o X - xoY = 8. First let c be even. Take 6 = +c, and Z so that
Then
Hence on continuing the process we have a solution with z2 I ab. Secondly let c be odd. Now impose the condition
aX
+ b Y + CZ E 0 (mod 2).
This defines the parity of 2. Since 6 is odd, the three expressions on the righthand side of (9) are divisible by 26, and so we now have (10) with 6 replaced by 26. Take 6 = c and Z with assigned parity so that
Then (1 1) is replaced by
If, I
2 -
This completes the proof.
<1+1
and
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
49
3. Quaternary quadratic equations'O. By means of a linear substitution, we may take the equation in the form .f(.z) =
alx:
+ a2xg + a3xz + a4xi = 0,
where the a's are integers. Theorem 6 The equation f(z) = 0 is sohable in rational integers i f and only i f f ( x ) = 0 (mod p') is soh>ablefor all primes p and,for ail r with (xl,x2, x3, x4,p) = 1, anrlj(.r) = 0 is soloable in real numbers. We may suppose that the a are square free since if a, = A1a? etc. where A , etc. are square free, a 1-1 correspondence is established between the solutions of
5
r=l
$ A , X ; = 0,
arx; =o,
r=l
by the relations
X,
=
alxl,. . .,
x1 = a2a3a4X1.. .
We may also suppose as in the case of the ternary quadratic that no three of the a's have a common factor. Since the coefficients a,, a2,a3,a4 do not all have the same sign, we may suppose that a, > 0, u4 > 0. Write g
=
+ a2x&
alx?
h
= -a3xz
- a4xi.
We shall prove the theorem by showing the existence of an integer a representable by both g and h. This will follow from the p-adic solvability of f ( z ) = 0. Let p,, p2,. . . ,p s be all the different odd primes dividing a1a2a3a4.Then for everyp from among pl, . . .,ps,and forp = 2 there is a p-adic solution of a14?
+ a& + a,(; + a,(:
=
0.
We may suppose in this solution (ti,(h, &, 6;) that none of the 6' are zero. For if = 0, the p-adic equation is solvable for arbitrary f4. This is obvious on replacing (,, t2,f 3 by t f i q-,, t & T ~t G , 73, where the 71 are arbitrary, and this gives a linear equation in t. Write
+
6, = a,[?
+
+
+ a24X = -a3[;
-
a45i.
We can choose the solution 5 so that b, # 0 in the p-adic field Q,. For if 6 , = 0, then both g and h represent any integer in Q, since g and h
50
DIOPHANTINE EQUATIONS
factorize in the p-adic field. We may also suppose that b, = 0 (mod p%) but not p A p + l, and also when p = 2. Consider the congruences with a > 0, a
= 6 , (mod
a
= b, (modpApP1), p
=pl, .
. . ,ps.
These determine a uniquely mod m = 2A2+3p:1+l. . .p;$+ Since b, f 0 (mod p A p P + l ) b,a-’ , = 1 (modp). Hence the quadratic character (b,a-l/p) = 1, and so b,a- is a quadratic residue in Qpand so is ap-adic square. Similarly b,a-’ = 1 (mod 8) and so b,a-l is a quadratic residue in Q2 and so is a 2-adic square. Since b, and a differ by a square factor in all the fields Q,, Q,, both the forms -ax:
+ a,X: + a,XZ,
- a x : - a3XE - a4Xi
represent zero in all the Q, and Q,. These forms have a zero in every p-adic field, except possibly when p I a. Also the forms are indefinite since a, > 0, a, < 0. Hence if a has only one prime factor q different from 2, p , , . . .,ps, the forms will also have a zero in QQ.Then from the ternary case of Theorem 1, we have
a = ale?
+ a2c&
a
=
-a3c32 - a&,
with rational c, and so finally a,cl
+ a2c: + a3cf + a4ci = 0.
The existence of such an a is proved by Dirichlet’s theorem that there are an infinity of primes in an arithmetical progression in which the first term is prime to the difference. Let a‘ be an arbitrary one of the a defined by the congruences above; write d = (a’, m). Then there exists an integer k 2 0 such that a’/d + km/d is a prime q. We take a = a’ km = qd. We now state without proof the conditions for the p-adic solvability of the equation
+
f ( x ) = alx:
+ a,x$ + a3xi + a 4 4 = 0,
wheref(x) is in the normal form, i.e. the a are square free integers and no three of the a have a common factor. The a of course must not all have the same sign. The p-adic equation is solvable for all p when (p, a,, a,, a3, a,) = 1, as is obvious from the result for the ternary quadratic. Next let p = p , , be an odd common prime factor of a, and a, for which ( -a3a4/p)= 1. Then we must have (( -a,a2p-,)/p) = 1. Similarly for the other prs. Finally 2-adic solvability is required. This is easily seen to be the solvability of f ( x ) = 0 (mod 8). The conditions are as follows. If a1a2a3a4= 1 (mod 8), then we require a, -k a, + a3 a, 3 0 (mod 8). There are no conditions if
+
ala2a3a4= 2, 3, 5, 6 , 7 (mod 8).
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
51
Suppose next that
= a, = O(mod2),
ala2a3a4= 4(mod 8) and a, say. Then if
~a,a,a3a4= I (mod 8),
we require +a,
+ j a , + a3 + a4 = j(agaz - 1) (mod 8).
No conditions are required if ~a,a,a,a, = 3, 5, 7 (mod 8).
We prove now the 4. Theorem 7 Every indefinite quadratic form j ( x ) in fiiie iiariables represents zero. It suffices to take f(z)= a1x:
+
'
..
+ a&,
where the a are square free, a, > 0, us < 0. Write g
=
alx:
+ a&,
h
= -a3xg - a4xi
- a&.
We argue as in the case of four variables. We find by Dirichlet's theorem a positive rational integer a which is represented by both g and h in the real field, and in all p-adic fields Q, with the exclusion perhaps of a field Q, where q is an odd prime number not dividing any of the coefficients a,, a2, . . . . Then g represents a by the theory of the ternary quadratic. Also since h represents zero in Q,, it represents a rationally by the theory of the quaternary quadratic. The theorem follows as for n = 4. It is now trivial that an indefinite form f ( x l , x2,. . ., x,) with n > 5 variables represents zero. It is also easily proved that solutions exist with XlX,. . . .Y, # 0. 5. A much more difficult question is to investigate the solvability of two simultaneous homogeneous quadratic equations f ( 2 ) = 0, g(z) = 0 in n variables. Mordell" has shown by an application of Meyer's theorem that subject to some simple natural conditions, for example, that Af(s) p g ( z ) is not a definite form for all real values of A, p not both zero, that non-trivial solutions, i.e. not all zeros, exist if n 2 13. Swinnerton-Dyer12 has improved this to n 3 11. It is conjectured that n = 9 is the best possible result, but some equations in 9 variables are such that four of the variables must be zero. An interesting question is whether the two simultaneous equations in five variables are solvable when they are solvable in every p-adic field includingp,.
+
52
DIOPHANTINE EQUATIONS
REFERENCES 1. H. Hasse. Uber die Darstellbarkeit von Zahlen durch quadratische Formen
im Korper der rationalen Zahlen. J. reinp angew. Math., 152 (1923), 129-148; also 205-224. 2. L. Holzer. Minimal solutions of diophantine equations. Can. J. Math., 11 (1950), 238-244. 3. L. J. Mordell. On the equation ax2 + by2 - cz2 = 0. Monatshefte fur Math., 55 (1951), 323-327. 4. T. Skolem. On the diophantine equation ax2 + bya + cz2 = 0 . Rendiconti di Matemuticu e delle sue applicazioni (9,11 (1952), 88-102. 5. L. J. Mordell. On the magnitude of the integer solutions of the equation ax2 + by2 + cz2 = 0 . J. number theory, I (1968), 1-3. 6. M. Kneser. Kleine Losungen der diophantischen Gleichung ax2 + bya = cz’. Abh. Math. Sem. Hamburg, 23 (1959), 163-173. 7. J. W. S. Cassels. Bounds for the least solution of homogeneous quadratic equations. Proc. Camb. Phil. Soc., 51 (1955), 262-264. Addendum, ibid. 52 (1956), 664. 8. H. Davenport. Note on a theorem of Cassels. Proc. Camb. Phil. SOC., 53 (1957), 539-540. Addendum, ibid. 52 (1956), 664. 9. B. J. Birch and H. Davenport. Quadratic equations in several variables. Proc. Camb. Phil. SOC., 54 (1958), 135-138. 10. Z. I. Borevich and I. R. Shafarevich. “Number-theory”, Chapter I. Academic Press, New York and London (1966). 1 1 . L. J. Mordell. Integer solutions of simultaneous quadratic equations. Abh. Math. Sem. Hamburg, 23 (1959), 124-143. 12. H. P. F. Swinnerton-Dyer. Rational zeros of two quadratic forms. Acta Arith., 9 (1964), 260-270.
Homogeneous Equations of the Second Degree 1. Let
be an indefinite quadratic form with integer coefficients and of determinant D = IIursll # 0. We discuss the solvability in integers of the equationf(f(m) = 0. Results for n = 3 were given by Legendre for the diagonal forms, i.e. a23 = a31 = a,, = 0, and various modifications of his proof have been given by other writers. Results for diagonal forms in four and five variables were found by Meyer nearly a hundred years after Legendre's work. The general case was considered by Minkowski and Hasse' culminating in the latter's result given below as Theorem 1 , and embodying a new and important principle, namely, that solvability of the equation f(x) = 0 for all local (i.e. p-adic) fields implied global solvability (i.e. solution in integers). Other results give estimates for the magnitude of a solution in terms of the coefficients. For n = 3 , the first sharp results are due to Holzer,. His method is rather advanced, and so further on we shall give the less sharp results due to Mordel13 which can be proved by elementary means. These were also found later by S k ~ l e mMordel15 .~ has since found a very elementary proof of Holzer's results also given further on. An estimate sometimes better than Holzer's and depending upon results in quadratic fields was given by M. Kneser ". Estimates for the magnitude of a solution of the general quadratic equation have been given by Cassels7, Davenport8, Birch and Davenport', but they have no place here. These results are of great importance in many investigations. Theorem 1 The equation f(x) = 0 is solvable in integers if and only if the congruence f(x) = 0 (modp') is solvable for all primes p and integers r 2 1 with (x1, x2,. . ., xn, P) = 1. Stated otherwise the conditions imposed uponf(s) mean that the equation f(5) = 0 is solvable in every p-adic field including of course the real field pa. The condition is obviously necessary. It is well known that by means of a linear substitution X, = c,,xr . . . c,,x,, (r = 1 , . . ., n),
+
+
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
where the c are integers and c,, # 0, that the solution of the equationf(.c) is equivalent to that of an equation F(%)
=
A,Xf
+
. ..
+ AnXZ = 0.
43 =
0
(2)
This is easily proved by induction. We may suppose that a,, # 0 by a linear substitution if need be. Then
=
aIIf(4
X ? f fib),
where Similarly = X;
h22f1(.7.)
where
+
S2(T),
2 b2rxr, etc. n
X,
=
r=2
It can be shown that the substitution does not affect p-adic solubility. 2. The case n = 2 is trivial and so we commence with n = 3 and write f ( x ,y , Z)
=
ax2
+ by2 + C Z ~= 0.
(3)
We now may suppose that a, 6, c have no common factor, and also that x, y , z have no common factor. We also suppose that a, b, c do not all have
the same sign and thatf(x, y , z ) 3 0 (mod p') is solvable for all primes p and integers r >, 1 with (x, y , z , p) = 1. We now reducef(x, y , z ) to a canonical form. Firstly, we may suppose that a, 6, c are square-free. For let a = U , C X ~b, = b,p2, c = c,y2 where a,, b,, c1 are square-free. We have now a 1-1 correspondence between the solutions of equation (3) and of
m,,z1)
=
y1,
alx? + b,Yf
+ clz? = 0
given by n,x, =
ax,
d.Y = pyx,,
cllyl
=
py,
fly
=
yay,,
d,z, dz
= yz,
= apz1,
(x1, y,, Zl) = 1,
( x , y , z ) = 1.
y , z ) = 0 is p-adically solvable so is f,(x, y , z ) = 0 and conClearly if ,/(s, versely. Next we may suppose in equation (3) that (6, c) = (c, a ) = (a, b) = 1. For let p be a prime factor of (b, c) and so b = pbl, c = p c l . Then ax2 = 0 (modp), and so ,Y = p x l . Hence with a , = pa, (3) becomes a,s:
+ b,y2 +
c,z2
=
0.
44
DIOPHANTINE EQUATIONS
Here alblcl = abc/p and so lalblcll< label. Clearly the process, applied to all the prime factors of (6, c), (c, a), (a, b), must terminate at a stage when (6, c) = (c, a) = (a, b) = 1. Further the final equation is solvable as a congruence for all p'. We now write equation (3) in the form
ax2 + by2 -
CZ'
=
0, u > 0, b > 0, c > 0.
(4)
Let p be an odd prime divisor of a. Then by2 - cz2 = 0 (mod a) is solvable and so as in Mordel13 and Skolem4, by2 - cz2 = b(y
+ kz)(y - kz) (modp),
f(x, Y , 4 = LlM1 (mod P), identically in x, y , z where L1,M 1are linear forms in x, y , z. Similarly if q, r are odd prime divisors of 6, c, respectively.
say, and so
f(x, y , 4 = L2M2 (mod 41, f(x, y , z ) = L3M3(mod r).
ax2 + by2 - cz2 = (ax + by
Also
+ C Z ) ~(mod 2).
Hence since abc is square-free, on taking into account the various p, q, r, we have by the Chinese remainder theorem, f ( x , y, z ) z L M (mod abc),
where L,M are linear forms. We show now that we can find integers x, y, z # 0, 0, 0, such that
ax2 + by2 - cz2 = 0 (mod abc),
and
0 < 1x1 <
dbc,
0 6 (yI <
d/ca,
0
< IzI
(5)
<
d/ab,
provided we exclude the trivial case a = b = c = 1. Here the numbers dbc, dG, dab are irrational. The number of sets of integers x,y, z with
o
Uxi, yi, 21) Hence and
L(xl Ixl - x21 <
Then for x
=
- x2, yl
dbc,
2 z2) ,
(mod abc).
- y2, z1 - z2) = 0 (mod abc), Iyl - y21 < dG, Izl - z21 < dab.
x1 - x2,etc. -abc < axa
and so
L(x2,~
+ by2 -
ax2 + by2 -
CZ'
< bbc,
cz2 = 0 or abc.
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
45
In the second case,
+ by2)(z2 + ab) = c(z2 + ab)2, a(xz + by)2 + b(yz - ax)2 = c(z2 + ab)2. a x 2 + bY2 = cz2. (ax2
or
i.e.
This proves the solvability of equation (4). The result gives an estimate for the magnitude of the solution, but a better one is given by Mordell’s Theorem 2 There exists a non-triaial solution with 1x1 ,<
lyl <
d S C ,
d%,
IZI
< 2dab.
(6)
This is not so precise as Holzer’s2 result 1x1 < 6, (yl < di,IzJ < dab, where the equality signs can be omitted unless two of a, b, c, are equal to one. It suffices for a proof of this to show that we can find a solution not (0, 0,O) of the congruence f ( x , y , z) = ax2
with Then
1x1
< dZC,
+ by2 - cz2 = 0 (mod 4abc), Jyl <
dZa,
IZI
< 2dab.
If’(x, y , z)l < 4abc, i.e. f(x, y , z)
=
0.
We do this by applying Theorem 3 of Chapter 6 to five linear congruences. Suppose first that a, b, c are all odd. Then there exist integers A , B, C such that bA2 - c
= 0 (mod a), - cB2 + a = 0 (mod b),
aC2
+ b = 0 (mod c).
Three of the congruences are given by y - Az
= 0 (mod a),
z - Bx
= 0 (mod b),
x - Cy
= 0 (mod c).
Then from the first, ax2 + by2 - cz2 = (bA2 - c)z2 = 0 (mod a). Similarly for mod 6, mod c. We can satisfyf(x, y , z) = 0 (mod 4) by taking two linear congruences each to mod 2. We cannot have both a - c G 2 (mod 4), b - c = 2 (mod 4) for then a = b = - c (mod 4), and sof(x, y , z ) = 0 (mod 4) is not solvable. We may suppose that h - c = 0 (mod 41, and then the two congruences are x
= 0 (mod 2),
y
= z (mod 2).
46
DIOPHANTINE EQUATIONS
It now suffices to take rl
=
d/2bc,
r2
=
.\/2ca,
r3 = 2 4 2 .
in Theorem 3 of Chapter 6. Suppose next that one of a, b, c is even, say a. We can now make f(x, y , z ) = 0 (mod 3abc) if x, y , z satisfy the three congruences: y - Az
= O(mod+a),
z - Bx
= O(modb),
x - Cy
= O(modc).
We now satisfy f(x, y , z ) = 0 (mod 8) by taking two linear congruences one to mod 2 and the other to mod 4. These are x where I, m satisfy
+
= 1y (mod 2),
z
=
m y (mod 4),
a12 + b - em2 = 0 (mod 8).
If a b - c = 0 (mod 8), we take I = m = 1. If b - c = 0 (mod 8), we take I = 0, m = 1. Then as before, Theorem 3 of Chapter 6 gives a solution. The equation is impossible when b - c = 4 or a (mod 8 ) as follows on taking congruences mod 8. This finishes the proof. To summarize, we have proved
Theorem 3 rfa, b, care square free, (a, b) = (b, c) the same sign, then the equation ax2
=
(c, a )
=
1 , anda, b, c do not have
+ by2 + cz2 = 0
has non-trivial integer solutions if and only if - bc is a quadratic residue of a, and so for every prime factor of a, i.e. x2 + bc = 0 (mod a) is solvable, and similarly for - ca and - ab, and if ax2 + by2 + cz2 = 0 (mod 8)
is solvable. These give N , say, necessary and sufficient conditions. It can be shown that if any N - 1 of these conditions are satisfied, so is the remaining one. This is a consequence of the laws of quadratic reciprocity and the supplementary laws, namely, that if p , q are two different odd primes
and
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
47
There is, however, no need here to go into details. The result is well known from Hilbert’s theory of the norm-residue symbol. Theorem 4 If one integer solution not (0, 0,O)of the irreducible equation
f ( x ,y , z )
=
ax2
+ by2 + cz2 + 2fyz + 2gzx + 2hxy = 0
exists, then the general solution with ( x ,y , z ) expressions of the form x = Alp2
=
(7) 1 is given by a j k i t e number of
+ Bipq + Clq2, y = A2p2 + B 2 ~ 4+ C2q2, z = A3p2 + B3pq + c3q2,
(8) where p , q take all integer values with ( p , q) = 1, and the A , B, C are integer constants. For let a solution (x,, yo, z,) be known. Write z = rz,, x = rx, p, y = ry, q, where p , q, r are rational numbers. Then we have a linear equation in r from which r = f(p, q, O)/L(p,q ) where L is a linear function of p and q. Hence ignoring denominators, we have with an appropriate factor 8,
+
+
8 . ~= f l ( P , q),
8Y = f X P , q), 8,z = f 3 ( P , q), where fi,f2,f3 are binary quadratics with integer coefficients, and we may now suppose that p , q are integers and that ( p , q ) = 1. A simple elimination procedure shows that the common factors of f l ,f 2 , f 3are independent of p , q. These and S can be removed on replacing p , q by a linear substitution on P. 4. The method of Theorem 4 has been used by Mordel15to give an elementary proof of Holzer’s
Theorem 5 The solvable equation ax2 + by2 + cz2 = 0 taken in the canonicalform with a > 0, b > 0, c < 0 has a non-trivial solution with 1x1 I
dm,
dW,
IyI I
IZI
I
dZ
It is easily seen that all the equality signs may be removed unless two of the a, b, c are unity. Let (x,, yo, z,) be a solution with (x,, yo, z,) = 1. We show that if for this solution, ( z o ( > dab, then another solution (x, y , z) can be found with IzI < lzol. Hence there exists a solution ( x , y , z ) with IzI I dab. The inequalities for x , y are now obvious. Put
x
+ tx,
+
z = zo y = y, tY, where X , Y, 2, are integers to be determined later. Then = x,
(ax2
+ tz,
+ b Y 2 + cZ2)t + 2(axoX + byoY + czoZ) = 0.
48
DIOPHANTINE EQUATIONS
1
Hence, on neglecting a denominator, a solution, say, (x, y , z), is given by 6~ = zo(aX2
ax Sy
= =
xo(aX2 yo(aX2
+ 6 Y 2 + cZ') - 2Z(axoX + by0 Y + CZOZ), + b Y2 + - 2X(axoX + by, Y + C Z O Z ) , + 6 Y 2 + cZ') - 2 Y(uxOX + by0 Y + CZOZ), CZ2)
(9)
where S is a common divisor of the three expressions on the right. Further x , y , z are integers if
Ale,
6l(Xyo
For from ax;
-
Yxo),
+ by; + cz; = 0,
it easily follows that (S, abxoyo) = 1. From (9) it suffices to show that Q = a X 2 + b Y 2 = O(mod6). P = a x o X + 6yoY = O(mod6), Since also
X
= xo Y/yo, Q
E
P
(ax;
=
Y(axi
+ by;)/yo = 0 (mod 8);
+ by:) Y z / y ; E 0 (mod 6).
From (9)
Suppose now that z i > ab. Take X , Y as a solution of y o X - xoY = 8. First let c be even. Take 6 = +c, and Z so that
Then
Hence on continuing the process we have a solution with z2 I ab. Secondly let c be odd. Now impose the condition
aX
+ b Y + CZ E 0 (mod 2).
This defines the parity of 2. Since 6 is odd, the three expressions on the righthand side of (9) are divisible by 26, and so we now have (10) with 6 replaced by 26. Take 6 = c and Z with assigned parity so that
Then (1 1) is replaced by
If, I
2 -
This completes the proof.
<1+1
and
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
49
3. Quaternary quadratic equations'O. By means of a linear substitution, we may take the equation in the form .f(.z) =
alx:
+ a2xg + a3xz + a4xi = 0,
where the a's are integers. Theorem 6 The equation f(z) = 0 is sohable in rational integers i f and only i f f ( x ) = 0 (mod p') is soh>ablefor all primes p and,for ail r with (xl,x2, x3, x4,p) = 1, anrlj(.r) = 0 is soloable in real numbers. We may suppose that the a are square free since if a, = A1a? etc. where A , etc. are square free, a 1-1 correspondence is established between the solutions of
5
r=l
$ A , X ; = 0,
arx; =o,
r=l
by the relations
X,
=
alxl,. . .,
x1 = a2a3a4X1.. .
We may also suppose as in the case of the ternary quadratic that no three of the a's have a common factor. Since the coefficients a,, a2,a3,a4 do not all have the same sign, we may suppose that a, > 0, u4 > 0. Write g
=
+ a2x&
alx?
h
= -a3xz
- a4xi.
We shall prove the theorem by showing the existence of an integer a representable by both g and h. This will follow from the p-adic solvability of f ( z ) = 0. Let p,, p2,. . . ,p s be all the different odd primes dividing a1a2a3a4.Then for everyp from among pl, . . .,ps,and forp = 2 there is a p-adic solution of a14?
+ a& + a,(; + a,(:
=
0.
We may suppose in this solution (ti,(h, &, 6;) that none of the 6' are zero. For if = 0, the p-adic equation is solvable for arbitrary f4. This is obvious on replacing (,, t2,f 3 by t f i q-,, t & T ~t G , 73, where the 71 are arbitrary, and this gives a linear equation in t. Write
+
6, = a,[?
+
+
+ a24X = -a3[;
-
a45i.
We can choose the solution 5 so that b, # 0 in the p-adic field Q,. For if 6 , = 0, then both g and h represent any integer in Q, since g and h
50
DIOPHANTINE EQUATIONS
factorize in the p-adic field. We may also suppose that b, = 0 (mod p%) but not p A p + l, and also when p = 2. Consider the congruences with a > 0, a
= 6 , (mod
a
= b, (modpApP1), p
=pl, .
. . ,ps.
These determine a uniquely mod m = 2A2+3p:1+l. . .p;$+ Since b, f 0 (mod p A p P + l ) b,a-’ , = 1 (modp). Hence the quadratic character (b,a-l/p) = 1, and so b,a- is a quadratic residue in Qpand so is ap-adic square. Similarly b,a-’ = 1 (mod 8) and so b,a-l is a quadratic residue in Q2 and so is a 2-adic square. Since b, and a differ by a square factor in all the fields Q,, Q,, both the forms -ax:
+ a,X: + a,XZ,
- a x : - a3XE - a4Xi
represent zero in all the Q, and Q,. These forms have a zero in every p-adic field, except possibly when p I a. Also the forms are indefinite since a, > 0, a, < 0. Hence if a has only one prime factor q different from 2, p , , . . .,ps, the forms will also have a zero in QQ.Then from the ternary case of Theorem 1, we have
a = ale?
+ a2c&
a
=
-a3c32 - a&,
with rational c, and so finally a,cl
+ a2c: + a3cf + a4ci = 0.
The existence of such an a is proved by Dirichlet’s theorem that there are an infinity of primes in an arithmetical progression in which the first term is prime to the difference. Let a‘ be an arbitrary one of the a defined by the congruences above; write d = (a’, m). Then there exists an integer k 2 0 such that a’/d + km/d is a prime q. We take a = a’ km = qd. We now state without proof the conditions for the p-adic solvability of the equation
+
f ( x ) = alx:
+ a,x$ + a3xi + a 4 4 = 0,
wheref(x) is in the normal form, i.e. the a are square free integers and no three of the a have a common factor. The a of course must not all have the same sign. The p-adic equation is solvable for all p when (p, a,, a,, a3, a,) = 1, as is obvious from the result for the ternary quadratic. Next let p = p , , be an odd common prime factor of a, and a, for which ( -a3a4/p)= 1. Then we must have (( -a,a2p-,)/p) = 1. Similarly for the other prs. Finally 2-adic solvability is required. This is easily seen to be the solvability of f ( x ) = 0 (mod 8). The conditions are as follows. If a1a2a3a4= 1 (mod 8), then we require a, -k a, + a3 a, 3 0 (mod 8). There are no conditions if
+
ala2a3a4= 2, 3, 5, 6 , 7 (mod 8).
7.
HOMOGENEOUS EQUATIONS OF THE SECOND DEGREE
51
Suppose next that
= a, = O(mod2),
ala2a3a4= 4(mod 8) and a, say. Then if
~a,a,a3a4= I (mod 8),
we require +a,
+ j a , + a3 + a4 = j(agaz - 1) (mod 8).
No conditions are required if ~a,a,a,a, = 3, 5, 7 (mod 8).
We prove now the 4. Theorem 7 Every indefinite quadratic form j ( x ) in fiiie iiariables represents zero. It suffices to take f(z)= a1x:
+
'
..
+ a&,
where the a are square free, a, > 0, us < 0. Write g
=
alx:
+ a&,
h
= -a3xg - a4xi
- a&.
We argue as in the case of four variables. We find by Dirichlet's theorem a positive rational integer a which is represented by both g and h in the real field, and in all p-adic fields Q, with the exclusion perhaps of a field Q, where q is an odd prime number not dividing any of the coefficients a,, a2, . . . . Then g represents a by the theory of the ternary quadratic. Also since h represents zero in Q,, it represents a rationally by the theory of the quaternary quadratic. The theorem follows as for n = 4. It is now trivial that an indefinite form f ( x l , x2,. . ., x,) with n > 5 variables represents zero. It is also easily proved that solutions exist with XlX,. . . .Y, # 0. 5. A much more difficult question is to investigate the solvability of two simultaneous homogeneous quadratic equations f ( 2 ) = 0, g(z) = 0 in n variables. Mordell" has shown by an application of Meyer's theorem that subject to some simple natural conditions, for example, that Af(s) p g ( z ) is not a definite form for all real values of A, p not both zero, that non-trivial solutions, i.e. not all zeros, exist if n 2 13. Swinnerton-Dyer12 has improved this to n 3 11. It is conjectured that n = 9 is the best possible result, but some equations in 9 variables are such that four of the variables must be zero. An interesting question is whether the two simultaneous equations in five variables are solvable when they are solvable in every p-adic field includingp,.
+
52
DIOPHANTINE EQUATIONS
REFERENCES 1. H. Hasse. Uber die Darstellbarkeit von Zahlen durch quadratische Formen
im Korper der rationalen Zahlen. J. reinp angew. Math., 152 (1923), 129-148; also 205-224. 2. L. Holzer. Minimal solutions of diophantine equations. Can. J. Math., 11 (1950), 238-244. 3. L. J. Mordell. On the equation ax2 + by2 - cz2 = 0. Monatshefte fur Math., 55 (1951), 323-327. 4. T. Skolem. On the diophantine equation ax2 + bya + cz2 = 0 . Rendiconti di Matemuticu e delle sue applicazioni (9,11 (1952), 88-102. 5. L. J. Mordell. On the magnitude of the integer solutions of the equation ax2 + by2 + cz2 = 0 . J. number theory, I (1968), 1-3. 6. M. Kneser. Kleine Losungen der diophantischen Gleichung ax2 + bya = cz’. Abh. Math. Sem. Hamburg, 23 (1959), 163-173. 7. J. W. S. Cassels. Bounds for the least solution of homogeneous quadratic equations. Proc. Camb. Phil. Soc., 51 (1955), 262-264. Addendum, ibid. 52 (1956), 664. 8. H. Davenport. Note on a theorem of Cassels. Proc. Camb. Phil. SOC., 53 (1957), 539-540. Addendum, ibid. 52 (1956), 664. 9. B. J. Birch and H. Davenport. Quadratic equations in several variables. Proc. Camb. Phil. SOC., 54 (1958), 135-138. 10. Z. I. Borevich and I. R. Shafarevich. “Number-theory”, Chapter I. Academic Press, New York and London (1966). 1 1 . L. J. Mordell. Integer solutions of simultaneous quadratic equations. Abh. Math. Sem. Hamburg, 23 (1959), 124-143. 12. H. P. F. Swinnerton-Dyer. Rational zeros of two quadratic forms. Acta Arith., 9 (1964), 260-270.